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Complex Numbers question

2020 · 3 Sep · Shift 2 · Q37
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Complex Numbers question

2020 · 3 Sep · Shift 2 · Q37

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z1 , z2 are complex numbers such that Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and arg(z1 - z2) = π6{\pi \over 6}6π​, then Im(z1 + z2 ) is equal to :
  1. A
    32{{\sqrt 3 } \over 2}23​​
  2. B
    13{1 \over {\sqrt 3 }}3​1​
  3. C
    23{2 \over {\sqrt 3 }}3​2​
  4. D
    23{2\sqrt 3 }23​
View written solutionFree

Correct answer: D

Let

\qquad z_2=x_2+iy_2$$ with $x_1,x_2\in\mathbb R$ and $y_1,y_2\in\mathbb R$. We are given: $$\operatorname{Re}(z_1)=|z_1-1|, \qquad \operatorname{Re}(z_2)=|z_2-1|$$ and $$\arg(z_1-z_2)=\frac{\pi}{6}.$$ We need to find $$\operatorname{Im}(z_1+z_2)=y_1+y_2.$$ --- ## 1. Use the condition $\operatorname{Re}(z)=|z-1|$ Take a general complex number $$z=x+iy.$$ Then $$\operatorname{Re}(z)=x$$ and $$|z-1|=|(x-1)+iy|=\sqrt{(x-1)^2+y^2}.$$ Given $$x=\sqrt{(x-1)^2+y^2}.$$ Since modulus is non-negative, we must have $x\ge 0$. Now square both sides: $$x^2=(x-1)^2+y^2.$$ Expand: $$x^2=x^2-2x+1+y^2.$$ So, $$2x=1+y^2$$ which gives $$x=\frac{1+y^2}{2}.$$ Thus every such complex number satisfies $$z=\frac{1+y^2}{2}+iy.$$ So for $z_1,z_2$, $$x_1=\frac{1+y_1^2}{2}, \qquad x_2=\frac{1+y_2^2}{2}.$$ --- ## 2. Use the argument condition Now $$z_1-z_2=(x_1-x_2)+i(y_1-y_2).$$ Since $$\arg(z_1-z_2)=\frac{\pi}{6},$$ we have $$\frac{y_1-y_2}{x_1-x_2}=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}},$$ with $x_1-x_2>0$. Now compute $x_1-x_2$: $$x_1-x_2=\frac{1+y_1^2}{2}-\frac{1+y_2^2}{2}=\frac{y_1^2-y_2^2}{2}$$ $$=\frac{(y_1-y_2)(y_1+y_2)}{2}.$$ Hence $$z_1-z_2=\frac{(y_1-y_2)(y_1+y_2)}{2}+i(y_1-y_2).$$ Factor out $(y_1-y_2)$: $$z_1-z_2=(y_1-y_2)\left(\frac{y_1+y_2}{2}+i\right).$$ So $$\tan\arg(z_1-z_2)=\frac{y_1-y_2}{\frac{(y_1-y_2)(y_1+y_2)}{2}}=\frac{2}{y_1+y_2},$$ provided $y_1\ne y_2$ (which must hold, otherwise argument is undefined). Since $$\arg(z_1-z_2)=\frac{\pi}{6},$$ we get $$\frac{2}{y_1+y_2}=\frac{1}{\sqrt{3}}.$$ Therefore, $$y_1+y_2=2\sqrt{3}.$$ Hence, $$\operatorname{Im}(z_1+z_2)=y_1+y_2=2\sqrt{3}.$$ --- ## 3. Check the options The value is $$2\sqrt{3}.$$ So the correct option is: **D: $2\sqrt{3}$** --- ## 4. Compare with stored correct answer Stored correct answer: **D** Our derived answer: **D** They match.
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